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Welcome to the Lesson 6 of the course on Strength
of Materials. In this particular lesson
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we are going to discuss certain aspects of
analysis of stress.
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..
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Once this particular lesson is completed one
should be able to understand the concept of
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the stresses in polar coordinate system, you
will be able to understand the concept of
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the
stress for axi-symmetric bodies which eventually
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can be derived from this polar
coordinate system of stresses. We will also
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look into how to evaluate stresses at different
points.
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.
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This particular lesson includes the recapitulation
of the lessons we discussed already such
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as evaluation of stresses in polar coordinate
system and examples for evaluation of
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stresses at particular point in the stress
body.
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..
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Some questions to be answered:
What happens to octahedral stresses when first
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invariant is 0?
Now let us look into octahedral stresses.
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The normal stresses on the octahedral planes
which we had calculated sigma octahedral is
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equal to 1 by 3 ( σ 1 plus σ 2 plus σ 3
). We had
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defined the octahedral planes as the planes
which are equally inclined with the principal
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axis reference system. And thereby the stresses
which are acting σ 1 , σ 2 and σ 3 in the
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rectangular stress system and the octahedral
stress as defined is 1 by 3 ( σ 1 plus σ
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2 plus σ 3 )
which is the summation of the normal stresses
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in three dimensional stress system, this is
called as first invariant.
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If you remember tau oct square is equal to
2 by 9( σ 1 plus σ 2 plus σ 3 ) whole square
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minus
6 by 9 ( σ 1 σ 2 plus σ 2 σ 3 plus σ
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3 σ 1 ). Hence as it has been asked if ( σ
1 plus σ 2 plus σ 3 ) is
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equal to 0 then eventually the normal stress
on octahedral plane, sigma octahedral is
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equal to 0. So, if the first invariant is
0 then the octahedral normal stress is equal
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to 0,
only shear stress will exist on the octahedral
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plane.
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..
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Now the second question which was posed was:
What is the value of the shear stress where
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maximum normal stress occurs?
The third question is:
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What the value of normal stress is where maximum
shear stress occurs?
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Probably these two questions can be answered
through the same diagram.
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If you remember, last time we had shown how
to plot the Mohr’s circle. This is the
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sigma-axis and this is the τ -axis. Now,
if we draw the Mohr’s circle of stress,
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the centre
of the Mohr’s circle from τ -axis is given
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as ( σ x plus σ y ) by 2 and the maximum
value of
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the normal stress at this point in this particular
plane we normally designate as σ 1 , and
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the
minimum value of the normal stress is σ 2
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. So, if you note in these two planes where
the
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maximum and minimum normal stress acts the
value of the shear stresses are 0. So the
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plane where the maximum normal stress acts
there the value of the shear stresses are
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0
and these planes are called as principal planes.
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..
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The third question was:
What is the value of the normal stress in
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the plane where the maximum shear stress acts?
These are the planes where the maximum shear
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stress acts. This is the maximum positive
shear and this is the maximum negative shear.
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If you note, the maximum shear the value
of the shear stress is that of the radius
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of the Mohr’s circle is τ max. If you note
here, in
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this plane we have the normal stress which
is equal to this particular magnitude
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( σ x plus σ y ) by 2. So, from this diagram
it self you can answer both the questions.
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The
planes where the normal stresses are at maximum
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the shear stresses are 0 and the planes
where shear stresses are maximum there normal
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stresses exist and the value of normal
stresses are ( σ x plus σ y ) by 2.
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..
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Let us look into aspects of how to evaluate
stresses in polar coordinates?
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So far we have discussed about the rectangular
stresses in a body where we have assumed
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that the boundaries are straight boundaries.
Now there are several cases where other than
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the straight boundaries we get problems, we
get structural elements where the surfaces
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are curved and to represent the stresses on
those curved surfaces.
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It is ideal to represent them in terms of
a coordinate system which can be expressed
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in
terms of radius and the rotational angle θ
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which we call as cylindrical axis or polar
reference axis. If we have cylindrical body
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of this particular form, in this we have earlier
seen the reference axis system as x, y, and
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z.
Now if we select a point here on this body,
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let us say this is P, the coordinate of this
particular point can be described by these
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coordinates x, y and z. Also, this particular
point can be represented through another reference
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system which if we project this point
on this xz-plane and draw a line over here
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and if we define this particular angle as
θ and
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its projected length as vector r and this
distance as y then the coordinate of this
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particular
point can be expressed as a function of r,
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θ , and y.
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This is with reference to the Cartesian
system x, y, and z and this particular reference
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is with reference to the polar coordinate
system which is r, θ and y.
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Now, if we look into the plan of this or the
cross section of this, then if we draw two
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radial lines from the centre, let us say that
this particular radial line is at an angle
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θ then
these two radial lines make a small angle
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of d θ . Now if we take a small element over
here and try to look into the stresses that
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will act then we will have two planes, this
particular plane normal to this plane is in
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the direction of the radius; we call this
as the rplane. Over these we will have the
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stresses which are acting as the normal stress
and so is
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this which we call as σ r .
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.The stress normal to this surface acting
along the circumferential direction is called
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as σ θ .
Also, we will have the tangential stress on
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this plane as well as on this plane and we
will
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have tangential stress in the radial direction
as well. This tangential stress which we
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defined as the shearing stresses is tau on
the r-plane acting in the direction of θ
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, we call
this as τ rθ .
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The other tangential stress on the θ -plane
is τ θ r , eventually τ rθ is equal to
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τ θ r . So we
define the state of stress on this particular
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body at a particular point is equal to σ
r the
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radial stress, the tangential stress is σ
θ , and the shearing stresses τ rθ . If
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a particular body
at a point is subjected to these stresses
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then how we arrive at the equations for
equilibrium.
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.
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Here I have tried to represent state of stress
at a particular point in a stress body, and
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as
we have designated that this particular stress
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which is acting normal to the r-plane, let
us
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call that as σ r . Let us assume that this
particular distance is dr. The first radial
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line let us
call, this is at distance of θ , and the
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small angle made by these two radial line
is d θ . The
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radial distance from the origin to the first
part of the element, let us call that as r.
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Hence the stresses which are acting σ r is
the normal stress, the tangential stress is
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τ rθ ,
and the circumferential normal stress as σ
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θ . So the stress which is acting on this
surface,
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since it is at a distance dr, while deriving
the equilibrium equations at the particular
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point
with reference to the rectangular Cartesian
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axis system if you try to find out the stress
at
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two different points then there is incremental
stresses which is σ r plus(∂ σ r by ∂r)dr.
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Likewise, the
tangential stress which is the shearing stress
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is equal to τ rθ plus (∂ τ rθ by
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.∂r)dr. On this we have σ θ , so on this
particular surface the normal circumferential
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stress
is equal to σ θ plus (∂ σ θ by ∂ θ
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)d θ .
The shearing stress which is acting on this
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surface is τ rθ is equal to τ rθ plus
(∂ τ rθ by
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∂ θ )d θ . Also, if we draw a tangent
at this particular point please note that
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normal stress
on this surface makes an angle of d θ by
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2. If we take the equilibrium of forces in
the
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radial direction and in the tangential direction
we can get the equations of equilibrium.
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Now on this particular plane, the area on
which this particular stress acts is r plus
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dr d θ ;
d θ being small this particular length you
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can write as r plus dr d θ and this particular
length as equal to rd θ .
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If we assume the thickness of this particular
element perpendicular to the plane of this
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board as unit then area of this particular
surface is equals to (r plus dr) d θ into
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1 and this
multiplied by the stress will give the force
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in this particular plane. Similarly the force
on
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this particular plane is σ r into rd θ into
1.
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Also, we have the tangential stresses on this
surface which are in the radial direction
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and
sigma θ will have the component in the tangential
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direction and also in the radial
direction and the component in the radial
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direction will be sigma θ (sin d θ by 2)
and d θ
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being small we can approximate sin d θ by
2 is equal to d θ by 2. So, if we write down
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the equations of equilibrium in the radial
direction then the equations we get are: σ
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r plus
(∂ σ r by ∂r) into dr is the normal stress
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on the outer plane multiplied by area which
is (r
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plus dr) d θ minus σ r rd θ (the normal
stress which is acting on the first plane
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where σ r
acts is equal to the σ r the stress times
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rd θ into 1) is the area so this is the force.
Then the shearing forces which are acting
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in the θ plane are: plus τ rθ plus (∂ τ
rθ by ∂ θ
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acting over the length d θ and the area is
dr minus τ rθ θ over the area dr(1) minus
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corresponding to the sigma θ we have σ θ
plus ∂ σ θ over a length d θ acting over
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the area
dr minus into d θ by 2, that sine component
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of that and this is minus so this is plus
or σ θ
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is acting hence this is plus and this is minus
so this is σ θ or this component called
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sine
component is acting in the reverse direction
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of the radial direction so this is also negative
and this is also negative so sigma θ dr d
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θ by 2 is equal to 0.
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..
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Hence on simplification this gives σ r rd
θ plus ∂ σ r r by dr into rd θ , when
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we multiply
with dr we have plus σ r dr d θ , now this
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particular term when multiplied by dr d θ
since
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we will have term dr square. And since this
being small we are neglecting that particular
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term, minus σ r rd θ plus τ rθ dr plus
∂ τ rθ by ∂ θ d θ dr minus τ rθ
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dr minus sigma θ dr d θ
by 2 minus d θ square. We neglect that minus
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σ θ dr d θ by 2. Now from this we find
that
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σ r rd θ and minus σ r rd θ these two
terms cancel out; τ rθ d θ and minus τ
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rθ d θ .
.
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.So we have ∂ σ r by ∂r dr rd θ plus
σ r dr d θ plus ∂ τ rθ d θ dr and minus
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σ θ by 2 and σ θ
by 2 if we combine them together this dr d
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θ and this is equal to 0 for the equilibrium
of
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the forces in the radial direction.
If we write it down and divide the whole by
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dr d θ then what is left out is, ∂ σ r
by ∂r plus
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1 by r ∂ τ rθ by ∂ θ plus ( σ r minus
σ θ ) by r is equal to 0. This is the equilibrium
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equation
in the radial direction. Similarly, if we
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take, equilibrium of the forces in the
circumferential direction then we can write
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down the equation as ( σ θ plus∂ σ θ
by ∂ θ
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d θ )dr minus σ θ dr plus ( τ rθ plus
∂ τ rθ by ∂r) (r plus dr) d θ minus
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τ rθ rd θ is equal to 0.
This gives the expression finally after simplification
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as, ∂ τ rθ by ∂r plus (1 by r) ∂ σ
θ by
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∂ θ plus τ rθ by r is equal to 0. So
these are the two equations, the equation
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∂ τ rθ by ∂r
plus (1 by r) ∂ σ θ by ∂ θ plus τ
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rθ by r is equal to 0 and ∂ σ r by ∂r
plus 1 by r ∂ τ rθ by ∂ θ
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plus ( σ r minus σ θ ) by r is equal to
0. These are represented in terms of the stresses
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as σ r ,
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τ rθ , and σ θ . They are written in terms
of the polar reference axis as θ and r.
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.
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These are the equations of equilibrium and
in this particular case we have not accounted
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for the components of the body forces. Both
in the radial and circumferential direction
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we
have neglected the body forces. Hence we have
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∂ σ r ∂r plus 1 by r ( τ rθ by ∂ θ
σ r ) plus
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( σ r minus σ θ ) by r is equal to 0; this
is the first equation of the equilibrium.
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∂ τ rθ by ∂r
plus 1 by r (∂ σ θ by ∂ θ ) plus τ
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rθ by r is equal to 0 is another equation
of equilibrium.
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These are the two equations of equilibrium
which are explained in reference to the polar
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coordinate system.
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..
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Thereby from this we can find out the stresses
that are referred in axi-symmetric body.
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We encounter several kinds of structural elements
where the stresses or the boundaries
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may not be perfectly straight; you can have
curved boundary over which there could be
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stress which can be radial stress or which
can be described by the stress σ r and σ
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θ and if
the loading on such body is symmetrical then
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we have a perfectly symmetrical body or
loading is perfectly symmetrical in its vertical
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direction.
Then if we take any cross section or any longitudinal
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section for that matter, if we take
section through the diameter at every section
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the level of the stress will be the same.
Hence it shows that the stresses at any of
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these sections are independent on wherever
we
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take the section, so the stresses are independent
of θ and these kinds of bodies are called
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as axi-symmetric bodies. That means these
bodies are perfectly symmetrical with
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reference to the vertical axis. For such bodies
if the loading also is vertical and
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symmetrical then any cross section we take,
at each section the same state of stress exists.
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This kind of stress and the body we call as
axi-symmetry bodies.
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The bodies which are perfectly symmetrical
referring to its vertical axis we call them
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as
axi-symmetry bodies and for axi-symmetry bodies
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if they are loaded symmetrically then
the stress components do not depend on θ
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. Therefore, any longitudinal section we take
the shear stress components are absent because
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we have the symmetric deformation and
thereby the shear stress components do not
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exist. If we take the absence of the shearing
stresses then the equilibrium equation reduces
209
00:27:46,880 --> 00:27:52,720
to ∂ σ r by ∂r plus ( σ r minus σ θ
) by r is
210
00:27:52,720 --> 00:28:08,549
equal to 0 where only the normal stresses
exist and the shearing stresses are absent.
211
00:28:08,549 --> 00:28:09,549
..
212
00:28:09,549 --> 00:28:15,980
We have tried to give an outline of the state
of stress if we refer them in terms of the
213
00:28:15,980 --> 00:28:22,720
polar coordinate system. Earlier in a stress
body we have looked into that, if we have
214
00:28:22,720 --> 00:28:25,620
the
rectangular components of the stresses σ
215
00:28:25,620 --> 00:28:29,980
x , σ y , and τ xy we looked into how to
evaluate the
216
00:28:29,980 --> 00:28:34,520
stresses at different points and on different
planes.
217
00:28:34,520 --> 00:28:42,419
Now, if we try to represent the stresses on
any plane in a polar coordinate system where
218
00:28:42,419 --> 00:28:48,679
the normal stresses σ r , σ θ , and shearing
stress τ rθ exist we have seen how to write
219
00:28:48,679 --> 00:29:09,269
down
the equations of equilibrium.
220
00:29:09,269 --> 00:29:23,149
Here if you look in this particular point
in the stress body the stresses given are;
221
00:29:23,149 --> 00:29:29,049
on a
horizontal plane the normal stress is 60 MPa,
222
00:29:29,049 --> 00:29:35,940
on a particular plane which is inclined with
reference to this horizontal plane is 45 degrees,
223
00:29:35,940 --> 00:29:42,850
the normal stresses are 50 MPa; the
shearing stress is 40 MPa; and on this horizontal
224
00:29:42,850 --> 00:29:52,340
plane we have shearing stress as 30
MPa. What we will have to compute is the normal
225
00:29:52,340 --> 00:30:02,169
and shear stresses which are acting on
the vertical plane using transformation equations.
226
00:30:02,169 --> 00:30:15,260
Here the given values are σ y which equals
to 60 MPa which is positive; τ xy is given
227
00:30:15,260 --> 00:30:23,350
as
30 MPa; and on this particular plane on which
228
00:30:23,350 --> 00:30:32,269
we have defined σ x prime the normal plane
is equal to 50 MPa and the shearing stress
229
00:30:32,269 --> 00:30:39,611
τ x ' y ' is equal to 40 MPa. We will have
to
230
00:30:39,611 --> 00:30:48,630
compute what is the value of σ x which is
acting on the vertical plane and correspondingly
231
00:30:48,630 --> 00:31:01,570
what is the shearing stress tau. These are
the two values we have to evaluate.
232
00:31:01,570 --> 00:31:11,130
If you remember the transformation equations
on any plane which is σ x prime is equal
233
00:31:11,130 --> 00:31:18,409
to
( σ x plus σ y ) by 2 the normal stresses
234
00:31:18,409 --> 00:31:31,019
plus ( σ x minus σ y ) by 2 into cos 2 θ
plus τ xy sin
235
00:31:31,019 --> 00:31:42,090
2 θ . Now in this particular problem the
stress on the inclined plane is given as σ
236
00:31:42,090 --> 00:31:44,000
x prime
237
00:31:44,000 --> 00:31:58,110
.and τ x ' y ' and normal to this particular
plane is at an angle of 45 degrees. So θ
238
00:31:58,110 --> 00:32:07,340
here is 45
degrees thereby 2 θ is 90 degrees. Now σ
239
00:32:07,340 --> 00:32:22,049
x prime is given as 50, so 50 is equal to
( σ x plus σ y ) by 2, as σ y is equal
240
00:32:22,049 --> 00:32:34,649
to 60, so ( σ x plus 60) by 2 plus ( σ x
minus 60) by 2 into
241
00:32:34,649 --> 00:32:42,899
cos of 90 is (0) plus τ xy sin 90 which is
(1).
242
00:32:42,899 --> 00:32:56,620
The second equation is; τ x ' y ' is equal
to 40 is equal to minus ( σ x minus σ y
243
00:32:56,620 --> 00:33:05,190
) by 2 cos 2 θ ,
where ( σ x minus σ y ) by 2 sin 2 θ (which
244
00:33:05,190 --> 00:33:12,100
is 1) plus τ xy cos 2 θ (which is equals
to 0). So
245
00:33:12,100 --> 00:33:32,010
from this we get minus σ x plus 60 is equal
to 80 or σ x is equal to minus 20 MPa. Now
246
00:33:32,010 --> 00:33:36,389
if
σ x is minus 20 if we substitute in this
247
00:33:36,389 --> 00:33:41,460
equation one for σ x value minus 20 plus
60 is
248
00:33:41,460 --> 00:33:51,130
equal to 40, 40 by 2 is equal to 20 so 50
is equal to 20 plus τ xy and thereby this
249
00:33:51,130 --> 00:33:52,130
gives you
250
00:33:52,130 --> 00:34:03,880
τ xy is 30 MPa, which is the shear stress
component in the horizontal plane. So, if
251
00:34:03,880 --> 00:34:09,020
we
draw the element now on which the stresses
252
00:34:09,020 --> 00:34:24,270
act we have on this as σ y , now we have
evaluated the σ x and also τ xy on this
253
00:34:24,270 --> 00:34:33,010
plane which is at an angle of 45 degrees with
the xplane on which the σ x prime and τ
254
00:34:33,010 --> 00:34:42,070
x ' y ' acts. These are the values of σ x
; σ x gives you
255
00:34:42,070 --> 00:34:47,480
minus 20.
Here if you see I have made a mistake that
256
00:34:47,480 --> 00:34:51,720
the stress is minus 20 so that indicates that
the
257
00:34:51,720 --> 00:34:57,710
normal stress will be acting in the opposite
direction, it will be a compressive stress
258
00:34:57,710 --> 00:35:05,750
whereas σ y is acting in the positive direction.
Here τ xy is positive so the direction of
259
00:35:05,750 --> 00:35:14,790
τ xy
is in the positive direction of y.
260
00:35:14,790 --> 00:35:25,520
That is the solution for this
particular problem.
261
00:35:25,520 --> 00:35:26,520
..
262
00:35:26,520 --> 00:35:43,390
Here is another problem. This particular problem
states that these are the stresses acting
263
00:35:43,390 --> 00:35:50,700
with the rectangular axis system which are
σ x , σ y , and τ xy and we will have to
264
00:35:50,700 --> 00:36:00,610
evaluate
the principal stresses, maximum shear stresses
265
00:36:00,610 --> 00:36:05,790
along with the normal stresses. We will
have to use Mohr’s circle to evaluate these
266
00:36:05,790 --> 00:36:15,830
quantities. Here if you look into the normal
stresses which is acting in the x-plane is
267
00:36:15,830 --> 00:36:19,250
compressive in nature having magnitude of
30
268
00:36:19,250 --> 00:36:34,570
MPa. So σ x is equal to minus 30 MPa, σ
y , the normal stress on the y-plane is 10
269
00:36:34,570 --> 00:36:35,871
MPa, so
270
00:36:35,871 --> 00:36:45,340
σ y is equal to plus 10 MPa. Then we have
the shearing stress.
271
00:36:45,340 --> 00:36:51,650
If you note the direction of the shearing
stress it is opposite to the positive y-direction.
272
00:36:51,650 --> 00:36:58,920
Also, this particular shear along with the
complimentary shear on the other face is
273
00:36:58,920 --> 00:37:06,270
causing rotation in the clockwise direction
which according to our sign convention is
274
00:37:06,270 --> 00:37:10,170
negative.
If you remember in the Mohr’s circle, we
275
00:37:10,170 --> 00:37:17,080
said that the shearing stresses which are
causing anticlockwise rotation is positive,
276
00:37:17,080 --> 00:37:20,660
since here this is the clockwise rotation
so
277
00:37:20,660 --> 00:37:28,140
these are negative shear. So, on this particular
surface, since this is causing anticlockwise,
278
00:37:28,140 --> 00:37:34,310
this particular shear is a positive shear.
Now if we try to represent these in the Mohr’s
279
00:37:34,310 --> 00:37:41,880
circle then let us see how it looks like.
280
00:37:41,880 --> 00:37:42,880
..
281
00:37:42,880 --> 00:37:52,500
This is the reference axis system and this
is the sigma plus and this is minus sigma,
282
00:37:52,500 --> 00:37:59,090
this is
plus tau and this direction is minus τ . Now
283
00:37:59,090 --> 00:38:08,610
on the x-plane we have σ x is equal to minus
30MPa which is this direction and τ xy is
284
00:38:08,610 --> 00:38:11,880
equal to minus 20 so this is the shear, so
we get
285
00:38:11,880 --> 00:38:18,150
the point somewhere here, which is σ x is
equal to minus 30 and τ is equal to minus
286
00:38:18,150 --> 00:38:23,370
20.
Then we have in the perpendicular plane the
287
00:38:23,370 --> 00:38:28,800
y-plane which is 90 degrees with reference
to the physical plane, here in the Mohr’s
288
00:38:28,800 --> 00:38:34,250
plane it will be 180 degrees and we have plus
σ y
289
00:38:34,250 --> 00:38:41,360
is equal to 10 MPa, and we have plus tau is
equal to 20 MPa.
290
00:38:41,360 --> 00:38:52,660
Now if we join these two points this is where
it crosses the sigma line, we get the centre
291
00:38:52,660 --> 00:39:01,970
of the Mohr’s circle. So with this as centre
O as the centre and OA as the radius we draw
292
00:39:01,970 --> 00:39:19,400
the circle. This gives us the Mohr’s circle
of which the centre is this particular point
293
00:39:19,400 --> 00:39:22,490
and
as you know the distance of the centre form
294
00:39:22,490 --> 00:39:33,380
the tau -axis ( σ x plus σ y ) by 2.
Now here σ x is equal to minus 30, σ y is
295
00:39:33,380 --> 00:39:41,220
equal to plus 10 that divided by 2, that gives
you minus 10. So this distance is minus 10
296
00:39:41,220 --> 00:39:44,870
MPa, this particular point refers to σ x
which
297
00:39:44,870 --> 00:39:59,400
is from here is minus 30, so the distance
from here to here is 20 which is ( σ x minus
298
00:39:59,400 --> 00:40:03,350
σ y )
by 2. As we have evaluated earlier this particular
299
00:40:03,350 --> 00:40:10,360
distance OO’is equal to ( σ x minus σ
y )
300
00:40:10,360 --> 00:40:18,960
by 2 and σ x is equal to minus 30 σ y is
equal to minus 10 by 2 is equal to minus 20.
301
00:40:18,960 --> 00:40:23,390
So
from here to here is 20, and this is τ xy
302
00:40:23,390 --> 00:40:43,320
which is also 20. So the radius or this particular
distance OA is equal to 202 + 202 is equal
303
00:40:43,320 --> 00:41:09,320
to 400 + 400 and this gives us the value of
28.28; 28.28 is the distance OA. Now, we are
304
00:41:09,320 --> 00:41:13,430
going to evaluate the principal stresses.
The
305
00:41:13,430 --> 00:41:23,470
principal stresses are: this is the maximum
principal stress which is σ 1 , and this
306
00:41:23,470 --> 00:41:31,420
is the
minimum principal stress, which is σ 2 . What
307
00:41:31,420 --> 00:41:33,890
will be the value of maximum principal
308
00:41:33,890 --> 00:41:45,150
.stress, which is acting on this particular
plane? Is the distance from here to here which
309
00:41:45,150 --> 00:41:50,480
is
in terms of the Mohr’s circle, is the radius
310
00:41:50,480 --> 00:42:00,650
minus this. So here the radius is 28.28 MPa
and this is the 10 MPa. So this is going to
311
00:42:00,650 --> 00:42:18,330
be equal to 18.2 MPa. So σ 1 is equal to
( σ x plus σ y ) by 2 plus the radius which
312
00:42:18,330 --> 00:42:25,730
is square root of ( σ x minus σ y ) by 2)
whole
313
00:42:25,730 --> 00:42:29,390
square plus τ xy square.
.
314
00:42:29,390 --> 00:42:35,961
This particular quantity is nothing but the
radius, which is equals to 28.28. So the value
315
00:42:35,961 --> 00:42:44,890
of σ 1 is equal to ( σ x plus σ y ) by
2 is equal to minus 10, and this is 28.28
316
00:42:44,890 --> 00:42:51,700
is equal to
minus 10 plus this is 28.28. Therefore, Stress
317
00:42:51,700 --> 00:43:01,590
is equal to 18.28 MPa. This is the
maximum principal stress. What will be the
318
00:43:01,590 --> 00:43:05,130
minimum principal stress? Minimum
principal stress will act on this particular
319
00:43:05,130 --> 00:43:10,960
plane, which is σ 2 , which is ( σ x plus
σ y ) by 2
320
00:43:10,960 --> 00:43:16,340
minus square root of (( σ x minus σ y ) by
2) whole square plus τ xy square.
321
00:43:16,340 --> 00:43:22,250
In terms of the Mohr’s circle, if you look
into, the distance is this radius plus this
322
00:43:22,250 --> 00:43:26,740
distance
and this radius we have got as 28.28 plus
323
00:43:26,740 --> 00:43:30,170
we have 10. So the distance from here to here
is
324
00:43:30,170 --> 00:43:37,430
38.28 and in terms of this equation, ( σ
x plus σ y ) by 2 is equal to minus 10 and
325
00:43:37,430 --> 00:43:42,860
this radius
is 28.28 which is negative. So in a combined
326
00:43:42,860 --> 00:43:55,350
form σ 2 is equal to minimum principal
stress is equal to minus 38.28 MPa.
327
00:43:55,350 --> 00:44:03,450
From this particular diagram, if we look into
that, this is the value of the radius and
328
00:44:03,450 --> 00:44:06,930
this is
the plane, where we get the value of maximum
329
00:44:06,930 --> 00:44:10,940
value of the shear stress and this is the
plane where you get the minimum value of the
330
00:44:10,940 --> 00:44:19,890
shear stress. Now at this particular point,
the value of the normal stress is this which
331
00:44:19,890 --> 00:44:24,360
is ( σ x plus σ y ) by 2 is equal to minus
10 MPa
332
00:44:24,360 --> 00:44:25,670
in this particular problem.
333
00:44:25,670 --> 00:44:34,200
.Therefore the maximum value of shear stress
is equal to radius is equal to 28.28 MPa.
334
00:44:34,200 --> 00:44:45,650
So
τ max is equal to 28.28 MPa and the corresponding
335
00:44:45,650 --> 00:44:52,700
normal stress on these particular planes
where the maximum and minimum share stress
336
00:44:52,700 --> 00:44:56,910
occurs, that is equals to the ( σ x plus
σ y )
337
00:44:56,910 --> 00:45:09,480
by 2 is equal to minus 10. So the normal stress
on this plane at is equals to ( σ x plus
338
00:45:09,480 --> 00:45:16,020
σ y )
by 2 is equal to minus 10 MPa. So these are
339
00:45:16,020 --> 00:45:27,020
the values which we get, they are: the
maximum principal stress σ 1 is equal to
340
00:45:27,020 --> 00:45:29,710
28.28 MPa; the minimum principal stress σ
2 is
341
00:45:29,710 --> 00:45:41,180
is equal to minus 38.28 MPa; the maximum shear
stresses τ max is equal to 28.28 MPa; the
342
00:45:41,180 --> 00:45:47,800
normal stress which acts on the maximum and
the minimum shear stress is equals to
343
00:45:47,800 --> 00:45:53,160
minus 10 MPa.
.
344
00:45:53,160 --> 00:45:58,690
We can evaluate the position of these planes,
the maximum, and minimum normal
345
00:45:58,690 --> 00:46:04,600
stresses with reference to the plane. Now
this is the plane which is representing the
346
00:46:04,600 --> 00:46:11,850
vertical plane, normal to the plane which
coincides with the x-axis, which you call
347
00:46:11,850 --> 00:46:16,240
as xplane.
Now, with reference to this particular plane,
348
00:46:16,240 --> 00:46:24,020
if we go in the anticlockwise direction this
particular angle will give us the value half
349
00:46:24,020 --> 00:46:29,300
of which is the orientation in the physical
plane which locates the maximum principal
350
00:46:29,300 --> 00:46:42,080
stress and perpendicular to that is the plane
where in this minimum normal stress acts.
351
00:46:42,080 --> 00:46:49,250
These are the values and that is how we can
compute the stresses using Mohr’s circle.
352
00:46:49,250 --> 00:46:50,250
..
353
00:46:50,250 --> 00:47:06,480
Here is another problem.
We already know that the octahedral stresses
354
00:47:06,480 --> 00:47:13,900
are the stresses which acts on the octahedral
plane and octahedral plane are the planes
355
00:47:13,900 --> 00:47:25,170
are equally inclined with respect to the principal
axes, σ 1 , σ 2 ,and σ 3 axes system. Now,
356
00:47:25,170 --> 00:47:28,300
if we know the values of principal stresses
at a
357
00:47:28,300 --> 00:47:35,780
point, we can compute the values of octahedral
stresses. Now, σ 1 , the maximum principal
358
00:47:35,780 --> 00:47:37,900
stresses is given as minus 2 plus
359
00:47:37,900 --> 00:47:45,150
10 MPa, σ 2 is equal to 1 and σ 3 is equal
to minus 2
360
00:47:45,150 --> 00:47:53,240
minus 10 MPa. And we will have to evaluate,
the values of octahedral stresses.
361
00:47:53,240 --> 00:48:09,410
So let us compute these values. So we have
σ 1 is equal to minus 2 plus 10 MPa, as the
362
00:48:09,410 --> 00:48:20,410
maximum principal stress. Then we have σ
2 , the second principal stress as 1 MPa σ
363
00:48:20,410 --> 00:48:27,370
3 as
equals to minus 2 minus 10 MPa. Now as we
364
00:48:27,370 --> 00:48:34,800
have seen in the beginning itself, that the
value of the sigma octahedral is equal to
365
00:48:34,800 --> 00:48:47,330
1 by 3 ( σ 1 plus σ 2 plus σ 3 ).
So here the values of σ 1 , σ 2 and σ 3
366
00:48:47,330 --> 00:48:54,450
are given, this is equal to 1 by 3(minus 2
plus 10 +
367
00:48:54,450 --> 00:49:03,580
1 minus 2 minus 10 ), 10 - these get cancelled
so minus 4 plus 1 is equal to minus 3 is
368
00:49:03,580 --> 00:49:14,270
equal to minus 1 MPa. This is the value of
σ of the normal octahedral stress.
369
00:49:14,270 --> 00:49:15,270
..
370
00:49:15,270 --> 00:49:22,960
For tau octahedral, we will compute in this
form, this is (tau octahedral) whole square
371
00:49:22,960 --> 00:49:29,869
is
equal to 2 by 9 ( σ 1 plus σ 2 plus σ 3
372
00:49:29,869 --> 00:49:45,880
) whole square minus 6 ( σ 1 σ 2 plus σ
2 σ 3 plus σ 3 σ 1 ).
373
00:49:45,880 --> 00:49:53,570
So here we have computed σ 1 plus σ 2 plus
σ 3 as minus 1, so this is minus 12 that
374
00:49:53,570 --> 00:50:00,730
is 2 by
9 (minus 6) of substitute the values of σ
375
00:50:00,730 --> 00:50:11,220
1 , σ 2 and σ 3 and this gives you a as
6 into σ 1 is
376
00:50:11,220 --> 00:50:26,570
minus 2 plus 10 σ 2 σ 3 is minus 2 minus
10 n and σ 1 and σ 3 is equal to plus 4
377
00:50:26,570 --> 00:50:41,190
minus
10. This gives you minus 10; so the τ octahedral
378
00:50:41,190 --> 00:50:50,670
is equal to
So this is the value of sigma octahedral and
379
00:50:50,670 --> 00:50:51,710
tau octahedral.
380
00:50:51,710 --> 00:50:56,440
78
is equal to 2.944 MPa.
381
00:50:56,440 --> 00:50:57,440
3
382
00:50:57,440 --> 00:50:58,440
..
383
00:50:58,440 --> 00:51:04,460
Here is another problem, this is the supporting
structure which supports a billboard, many
384
00:51:04,460 --> 00:51:11,270
a times we use boards, sign boards on which
the advertisements are put and these boards
385
00:51:11,270 --> 00:51:18,510
are supported by some steel structures and
these boards are subjected to the wind
386
00:51:18,510 --> 00:51:22,200
pressure.
The wind pressure when it comes on the board
387
00:51:22,200 --> 00:51:26,670
eventually it transfers the load on to the
supporting structure, hence this is one of
388
00:51:26,670 --> 00:51:33,070
the supporting structures in which we have
framework and all the members are connected
389
00:51:33,070 --> 00:51:38,230
in the pin joint and this is the force which
is acting on this member. Now our job is to
390
00:51:38,230 --> 00:51:46,450
find out the stress in each of these members
from this particular force. What we need to
391
00:51:46,450 --> 00:51:51,420
do is, first we evaluate forces in each of
the
392
00:51:51,420 --> 00:51:58,040
members which are arising from these external
forces, and thereby we can compute the
393
00:51:58,040 --> 00:52:03,730
stress since the cross sectional area of the
member is given, force divided by the area
394
00:52:03,730 --> 00:52:06,010
will
give us the stress.
395
00:52:06,010 --> 00:52:13,120
If we assume this angle as θ , then this
being 6 from the similar triangle we get this
396
00:52:13,120 --> 00:52:20,340
as 3,
so eventually this is also θ , and if we
397
00:52:20,340 --> 00:52:23,310
drop a perpendicular this is also θ , and
this is also
398
00:52:23,310 --> 00:52:32,720
θ and the values of cos θ is 4 by 5. This
particular hypotenuse, this being 3, and this
399
00:52:32,720 --> 00:52:43,790
being 4, is 5 so is this, this is also five,
so the values of cos θ is equal to 4 by 5
400
00:52:43,790 --> 00:52:48,950
and value
of sin θ is equal to 3 by 5. Now what we
401
00:52:48,950 --> 00:52:56,220
need to do is to draw free body diagram and
evaluate the forces.
402
00:52:56,220 --> 00:53:00,750
One section we can take here and draw the
free body diagram, and another section we
403
00:53:00,750 --> 00:53:07,860
can take here and draw the free body diagram
and we can compute the forces. Once we
404
00:53:07,860 --> 00:53:16,350
compute the forces, we can find out the stresses.
You try to compute the stresses for this
405
00:53:16,350 --> 00:53:37,740
particular problem; In this particular lesson
we tried to look into the stresses which we
406
00:53:37,740 --> 00:53:45,990
have evaluated earlier with reference to Cartesian
system σ x , σ y and τ xy . Also, we have
407
00:53:45,990 --> 00:53:53,140
tried to look into that if a body which is
does not have the straight boundary and there
408
00:53:53,140 --> 00:53:54,140
are
409
00:53:54,140 --> 00:53:59,370
.curved boundary, how to represent the stresses
σ r , σ θ and τ rθ which are in terms
410
00:53:59,370 --> 00:54:05,320
of polar
coordinate systems.
411
00:54:05,320 --> 00:54:06,320
.
412
00:54:06,320 --> 00:54:10,660
And also from there we have seen how to evaluate
stresses for the axi-symmetric bodies
413
00:54:10,660 --> 00:54:19,450
and also we looked in to some examples to
demonstrate how we an evaluate stresses at
414
00:54:19,450 --> 00:54:25,930
any point on the stress body either using
transformation equations or by the use of
415
00:54:25,930 --> 00:54:30,060
the
Mohr’s circle. We also tried to see how
416
00:54:30,060 --> 00:54:34,011
to compute octahedral stresses on a particular,
on
417
00:54:34,011 --> 00:54:40,740
a particular stress body; keep in mind this
octahedral stress is useful when we talk about
418
00:54:40,740 --> 00:54:47,930
the evaluation of the stress in the inelastic
when we go beyond the elastic strain.
419
00:54:47,930 --> 00:54:48,930
.
420
00:54:48,930 --> 00:54:58,450
.This particular lesson was last in the series
of the module stress analysis. We have
421
00:54:58,450 --> 00:55:11,940
computed or we have looked into the six lessons
in the particular module stress analysis.
422
00:55:11,940 --> 00:55:19,610
These six lessons if we look into chronologically,
in the first lesson I tried to give you the
423
00:55:19,610 --> 00:55:26,140
general concept of kinds of forces and what
really is the meaning of subject strength
424
00:55:26,140 --> 00:55:29,980
of
material, so it was introduced to you.
425
00:55:29,980 --> 00:55:30,980
.
426
00:55:30,980 --> 00:55:35,980
Subsequently the second lesson we had, I tried
to give you the concept of normal and
427
00:55:35,980 --> 00:55:45,180
shearing stresses and you know how to evaluate
the equations of the equilibrium from the
428
00:55:45,180 --> 00:55:51,560
stresses that are acing in the Cartesian system,
σ x , σ y and τ xy .
429
00:55:51,560 --> 00:55:58,600
In the third lesson we have tried to evaluate
stresses on any arbitrary plane, if we have
430
00:55:58,600 --> 00:56:04,350
any plane which is oriented with the value
of θ with reference to x-plane and thereby
431
00:56:04,350 --> 00:56:08,710
we
arrived at values of the stresses which we
432
00:56:08,710 --> 00:56:17,790
have defined as Cauchy’s stress formula.
We tried to evaluate the maximum normal stresses
433
00:56:17,790 --> 00:56:22,680
which we defined it as principal
stresses, and these principal stresses which
434
00:56:22,680 --> 00:56:26,800
we are acting in the principal planes, we
have
435
00:56:26,800 --> 00:56:33,910
tried to locate them in this particular lesson.
We have demonstrated that through few
436
00:56:33,910 --> 00:56:38,010
examples.
437
00:56:38,010 --> 00:56:39,010
..
438
00:56:39,010 --> 00:56:43,940
In the fourth lesson we have evaluated the
stress on any plane using transformation
439
00:56:43,940 --> 00:56:52,460
equations, in fact we have discussed this
lesson as well. How to evaluate stress any
440
00:56:52,460 --> 00:56:59,940
plane
using transformation equations or using Mohr’s
441
00:56:59,940 --> 00:57:05,480
circle?
In the fourth lesson we have discussed in
442
00:57:05,480 --> 00:57:09,340
detail how to construct a Mohr’s circle
if we
443
00:57:09,340 --> 00:57:18,710
know the stresses at a particular point in
a body. In the fifth lesson we tried to give
444
00:57:18,710 --> 00:57:24,480
you
the concept of octahedral stresses and also
445
00:57:24,480 --> 00:57:28,360
we looked into how to evaluate the stress
at a
446
00:57:28,360 --> 00:57:33,400
particular point in the stress body at different
planes.
447
00:57:33,400 --> 00:57:40,490
As you know the concept of the octahedral
stress, that this particular stress or the
448
00:57:40,490 --> 00:57:43,230
stresses
which act on the octahedral planes which are
449
00:57:43,230 --> 00:57:50,300
inclined equally with the principal axes
system and these stresses are useful when
450
00:57:50,300 --> 00:57:53,680
we talk about the evaluation of stress in
the
451
00:57:53,680 --> 00:57:58,550
inelastic stage.
452
00:57:58,550 --> 00:57:59,550
..
453
00:57:59,550 --> 00:58:05,570
In this particular lesson, we tried to give
you the concept of stresses in the polar
454
00:58:05,570 --> 00:58:11,390
coordinate systems that you have already looked
into, now having looked into these
455
00:58:11,390 --> 00:58:18,750
aspects of stresses here are some questions
to answer.
456
00:58:18,750 --> 00:58:19,750
.
457
00:58:19,750 --> 00:58:25,410
What are the equations of equilibrium in Cartesian
coordinate system of a stressed body?
458
00:58:25,410 --> 00:58:30,590
What are the equations of equilibrium in polar
coordinate system of a stressed body?
459
00:58:30,590 --> 00:58:38,150
What is the value of maximum shear stress
if σ 1 is equal to 10MPa, the maximum
460
00:58:38,150 --> 00:59:08,590
principal stress, and the minimum principal
stress σ 2 is equal to 0?
461
00:59:08,590 --> 00:59:08,590
.